Linear Programming Test

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Linear Programming Test - 30

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Question 1
1 / -0

Solve the following L.P.P. by graphical method:

Minimize: \(Z=6 x+2 y\) subject to \(x+2 y \geq 3, x+4 y \geq 4,3 x+y \geq 3, x \geq 0, y \geq 0\).

A
8

B
5

C
6

D
7

Solution

Given:

\(Z=6 x+2 y\)

\(x+2 y \geq 3\)

\( x+4 y \geq 4\)

\(3 x+y \geq 3\)

\( x \geq 0, y \geq 0\)

To draw the feasible region, construct table as follows:

Shaded portion XABCDY is the feasible region, whose vertices are \(\mathrm{A}(4,0), \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}(0,3)\).

\(B\) is the point of intersection of the lines \(x+4 y=4\) and \(x+2 y=3\)

Solving the above equations, we get:

\(\mathrm{x}=2, \mathrm{y}=\frac{1}{2} \)

\( \therefore \mathrm{B} =\left(2, \frac{1}{2}\right)\)

\(C\) is the point of intersection of the lines \(x+2 y=3\) and \(3 x+y=3\)

Solving the above equations, we get:

\(\mathrm{x}=\frac{3}{5}, y=\frac{6}{5} \)

\( \therefore C =\left(\frac{3}{5}, \frac{6}{5}\right)\)

Here, the objective function is \(Z=6 x+2 y\).

\(Z\) at \(A(4,0)=6(4)+2(0)=24\)

\(\mathrm{Z}\) at \(\mathrm{B}\left(2, \frac{1}{2}\right)=6(2)+2\left(\frac{1}{2}\right)=12+1=13\)

\(\mathrm{Z}\) at \(\left.C\left(\frac{3}{5}, \frac{6}{5}\right)=6\left(\frac{3}{5}\right)+2\left(\frac{6}{5}\right)=\frac{18}{5}\right)+\frac{12}{5}=6\)

\(\therefore Z\) at \(D(0,3)=6(0)+2(3)=6\)

\(\therefore Z\) has minimum value 6 at \(C\left(\frac{3}{5}, \frac{6}{5}\right)\) and \(D(0,3)\).

\(\therefore Z\) is minimum when, \(x=\left(\frac{3}{5}\right), y=\left(\frac{6}{5}\right), Z=6\) and \(x=0, y=3, Z=6\).
Question 2
1 / -0

An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

A
128000

B
136000

C
68000

D
124000

Solution

Let the airline sell \(x\) tickets of executive class and \(y\) tickets of economy class.
The mathematical formulation of the given problem is as follows:
Maximize \(\mathrm{Z}=1000 \mathrm{x}+600 \mathrm{y}\),
subject to the constraints,
\(x+y \leq 200\) ...(i)
\(x \geq 20\) ...(ii)
\(y-4 x \geq 0\) ...(iii)
\(x, y \geq 0\) ...(iv)
The feasible region determined by the constraints is as follows:

From equation (i) and (iii):
x + y\(\leq\) 200
x\(\leq\) 40
Put the value of x in equation (i):
y\(\leq\) 160
The corner points of the feasible region are \(A(20,80), B(40,160)\), and \(C(20,180)\).
The values of \(\mathrm{Z}\) at these corner points are as follows:
Corner Point Z= 1000x + 600y
A (20,80) 68000
B (40, 160) 136000
C (20, 180) 128000
The maximum value of \(\mathrm{Z}\) is 136000 at \((40,160)\).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is ₹ 136000 .
Question 3
1 / -0

Maximize \(Z=4 x+9 y\), subject to the constraints \(x =0\), y \(=0, x+5 y \leq 200,2 x+3 y \leq 134\).

A
359

B
856

C
382

D
756

Solution

\(Z=4 x+9 y\), subject to the constralnts \(x =0\), y \(=0, x+5 y \leq 200,2 x+3 y \leq 134\).

The corner points of feasible region are \(\mathrm{A}(10,38), \mathrm{B}(0,40), \mathrm{C}(0,0), \mathrm{D}(67,0)\).
The values of \(Z\) at the following polnts is
Corner
Points
\(Z=4 x+9 y\)
\(A(10,38)\) 382 Maximum
\(B(0,40)\) 360
\(C(0,0)\) 0
\(D(67,0)\) 268
The maxlmum value of \(Z\) is 382 at polnt \(A(10,38)\).
Question 4
1 / -0

The problem of maximizing \(z=x_{1}-x_{2}\) subject to constraints \(x_{1}~+~x_{2} \leq 10, ~x_{1} \geq 0, ~x_{2} \geq\) 0 and \(~x_{2} \leq 5\) has:

A
no solution

B
one solution

C
two solutions

D
more than two solutions

Solution

It is given that,\(z=x_{1}-x_{2}\) is subject to constraints:
\({x}_1+{x}_2 \leq 10\) ...(1)
\({x}_1 \geq 0\) ...(2)
\({x}_2 \geq 0\) ...(3)
\({x}_2 \leq 5\) ...(4)
Equations \(x_1=0\) \(x_2=0\)
\(x_1 + x_2 = 10\) \(x_2=10\) \(x_1=10\)
\(x_2 = 5\) - \(x_2 = 5\)

\(x_2 \) intersects \(x_1+x_2 = 10\) at B. From (1) and (4), coordinates of B are calculated as (5,5).

Points at Graph Coordinates \(z= x_1 - x_2\)
O (0,0) 0
A (10,0) 10
B (5,5) 0
C (0,5) -5

So, function \(z\) has feasible region in OABC and has maximum value at A(10,0).
Therefore, the function \(z\) has one solution.
Question 5
1 / -0

The minimum of the objective function \(\mathrm{Z}=2 x+10 y\) for linear constraints \(x-y \geq 0, x-5 y \leq-5, x \geq 0\) \(y \geq 0\), is:

A
15

B
16

C
17

D
18

Solution

Minimize \(Z=x+4 y\)
\(x+3 y \geq 3,2 x+y \geq 2, x \geq 0, y \geq 0\)
To draw the feasible region, construct table as follows:
\(\begin{array}{|l|c|c|} \hline \text{Inequality} & x+3 y \geq 3 & 2 x+y \geq 2 \\ \hline \text{Corresponding equation (of line)} & x+3 y=3 & 2 x+y=2 \\ \hline \text{Intersection of line with X-axis} & ((3,0) & (1,0) \\ \hline \text{Intersection of line with Y-axis} & (0,1) & (0,2) \\ \hline \text{Region} & \text{Non-origin side} & \text{Non-origin side} \\ \hline\end{array}\)
Shaded portion XABCY is the feasible region, whose vertices are \(A(3,0), B\) and \(C(0,2)\) \(B\) is the point of intersection of the lines \(2 x+y=2\) and \(x+3 y=3\)
\(\therefore B \equiv\left(\frac{3}{5}, \frac{4}{5}\right)\)
Here, the objective function is
\(Z=x+4 y \)
\(\therefore Z \text { at } A(3,0)=3+4(0) \)
\(=3\)
\(Z \text { at } B\left(\frac{3}{5}, \frac{4}{5}\right)=\frac{3}{5}+4\left(\frac{4}{5}\right) \)
\(=\frac{19}{5} \)
\(=3×8\)
\(Z\) at \(C(0,2)=0+4(2)\) \(=8\)
\(\therefore Z\) has minimum value 3 at \(x=3\) and \(y=0\).
Question 6
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Solve the following problem graphically:
Maximise \(Z =3 x+9 y\)
Subject to the constraints:
\(x+3 y \leq 60 \)
\(x+y \geq 10\)
\(x \leq y \)
\(x \geq 0, y \geq 0\)

A
\(180\)

B
\(150\)

C
\(110\)

D
\(60\)

Solution

Given:
Maximise \(Z =3 x+9 y\quad \cdots ~(1)\)
Subject to the constraints:
\(x+3 y \leq 60 \quad \cdots ~(2)\)
\(x+y \geq 10 \quad \cdots ~(3)\)
\(x \leq y \quad \cdots ~(4)\)
\(x \geq 0, y \geq 0\quad \cdots ~(5)\)
First of all, let us graph the feasible region of the system of linear inequalities \( (2)\) to \( (5)\). The feasible region ABCD is shown in the given figure. Note that the region is bounded. The coordinates of the corner points A, B, C and D are \( (0,10),(5,5),(15,15) \) and \( (0,20)\) respectively.
On evaluating intersection point C ( From equation \(2\) and \(4\) ):
\(x+3y=60\)
\(x=y\)
\(4y=60\)
\(\Rightarrow x=15\)
\(\Rightarrow y=15\)
Thus, the coordinates of C is \( (15,15) \).
On evaluating intersection point B (From equation \(3\) and \(4\) ):
\(x+y=10\)
\(x=y\)
\(2y=10\)
\(\Rightarrow y=5\)
\(\Rightarrow x=5\)
Thus, the coordinates of B is \( (5,5)\).
Corner Point Corresponding value of \(Z=3x=9y\)
\(A~(0, 10)\) \(90\)
\(B~(5,5)\) \(60\)
\(C~(15,15)\) \(180\)
\(D~(0,20)\) \(180\)
From the table, we find that the maximum value of \(Z\) on the feasible region occurs at the two corner points C \( (15,15)\) and D \( (0,20)\) and it is \(180\) in each case.
Question 7
1 / -0

Solve the linear programming problem by graphical method.
Maximize \(\mathrm{Z}=6 \mathrm{x}+3 \mathrm{y}\), subject to the constraints, \(4 x+y \geq 80, x+5 y \geq 115,3 x+2 y \leq 150, x, y \geq 0\).

A
252

B
254

C
285

D
298

Solution

Given:
Maximize \(\mathrm{Z}=6 \mathrm{x}+3 \mathrm{y}\), subject to the constraints,
\(4 x+y \geq 80\) ....(1)
\(x+5 y \geq 115\) ....(2)
\(3 x+2 y \leq 150\) ....(3)
\( x, y \geq 0\).
x=0 y=0
x+5y =115 y=23 x=115
4x+y =80 y=80 x=20
3x+2y =150 y=75 x=50

From eq (2) and (3), The coordinates of B are (40,15), from eq (1) and (3), the coordinates of C are (2,72) and from eq (1) and (2), the coordinates of A are (15,20).
The feasible region determined by the system of constraints is as follows:
The corner points of the feasible region are A \((15,20), B(40,15)\), and C\((2,72)\). The values of Z at these corner points are as follows.
Corner point Z=6x + 3y
A(15, 20) 150
B(40, 15) 285
C(2, 72) 228
Thus, the maximum value of Z is 285 at (40,15).
Question 8
1 / -0

Maximize: \(\mathrm{z}=3 \mathrm{x}+5 \mathrm{y}\)
Subject to: \(\mathrm{x}+4 \mathrm{y} \leq 24\)
\(3 \mathrm{x}+\mathrm{y} \leq 21 \)
\(\mathrm{x}+\mathrm{y} \leq 9 \)
\(\mathrm{x} \geq 0, \mathrm{y} \geq 0\)

A
\((\mathrm{x}, \mathrm{y})=\left(\frac{65}{11}, \frac{51}{61}\right), \mathrm{z}=\frac{435}{21}\)

B
\((\mathrm{x}, \mathrm{y})=\left(\frac{60}{11}, \frac{51}{11}\right), \mathrm{z}=\frac{435}{11}\)

C
\((\mathrm{x}, \mathrm{y})=\left(\frac{6}{11}, \frac{51}{6}\right), \mathrm{z}=\frac{5}{21}\)

D
\((\mathrm{x}, \mathrm{y})=\left(\frac{63}{121}, \frac{51}{6}\right), \mathrm{z}=\frac{95}{21}\)

Solution

The maximize: \(z=3 \mathrm{x}+5 \mathrm{y}\) Given subjective to,
\(x+4 y \leq 24 \)
\(3 x+y \leq 21 \)
\(x+y \leq 9, \quad x \geq 0, y \geq 0\)
Consider,
\(\mathrm{x}+4 \mathrm{y}=24 \).....(1)
\(3 \mathrm{x}+\mathrm{y}=21 \).....(2)
\(\mathrm{x}+\mathrm{y}=9\).....(3)
On solving equation (1) and (2), we get \(y=\frac{51}{11}\)
From equation (1),
\(\mathrm{x}=\frac{60}{11}\)
Therefore, \((\mathrm{x}, \mathrm{y})=\left(\frac{60}{11}, \frac{51}{11}\right)\) and \(z=3 x+5 y=\frac{435}{11}=39.54\)
Now,
On solving equation (2) and (3), we get \(\mathrm{x}=6\)
From equation (3),
\(y=3\)
Therefore, \((\mathrm{x}, \mathrm{y})=(6,3)\)
and \(z=3 x+5 y=33\)
Now,
On solving equation (1) and (3), we get \(\mathrm{y}=5\)
From equation (3),
\(\mathrm{x}=4\)
Therefore, \((\mathrm{x}, \mathrm{y})=(4,5)\)
and \(z=3 x+5 y=37\)
So, \((\mathrm{x}, \mathrm{y})=\left(\frac{60}{11}, \frac{51}{11}\right), \mathrm{z}=\frac{435}{11}\)
Question 9
1 / -0

Consider the Linear Programming function \(z= -3x+4y \) subjected to constraints,\(x+2 y \leq 8,~3 x+2 y \leq 12,~ x \geq 0, ~y \geq 0\), find the minimum value of \(z\).

A
-13

B
+13

C
-12

D
+12

Solution

Our problem is to minimize
\(z=-3 x+4 y\)
Subject to constraints,
\(x+2 y \leq 8\)
\( 3 x+2 y \leq 12 \)
\( x \leq 0, y \leq 0\)
From the table,
x=0 y=0
x=2y=8 y=4 x=8
3x+2y=12 y=6 x=4

\(\therefore\) Feasible region is \(\mathrm{OABCO}\).
\(x+2 y=8\)...(i)
\(3 x+2 y=\)12...(ii)
On solving equation (i) and (ii), we get
\(x=2\) and \(y=3\)
\(\therefore\) Intersection point is \(B\) and its coordinates is given by \((2,3)\).
The corner points of the feasible region are \(O(0,0), A(4,0), B(2,3)\) and \(C(0,4)\), The values of \(Z\) at these points are as follows:
Corner Point Z=−3x+4y
O (0,0) 0
A (4,0) -12⟶ Minimum
B (2,3) 6
C (0,4) 16
Therefore, the minimum value of \(Z\) is \(-12\) at the point \(A(4,0)\).
Question 10
1 / -0

Region represented by the in equation system \(x+y \leq 3, y \leq 6\) and \(x \geq 0, y \geq 0\) is:

A
Unbounded in first quadrant

B
Unbounded in first and second quadrant

C
Bounded in first quadrant

D
None of the these

Solution

Given:

\(x+y \leq 3, y \leq 6\) and \(x \geq 0, y \geq 0\)

Converting the given in equations into equations

\(x+y=3\).......(1)

\(y=6 \ldots \text {...(2) }\)

Reglon represented by \(x+y \leq 3\) :

The line \(x+y=3\) meets the coordinate axes are \(\mathrm{A}(3,0)\) and \(\mathrm{B}(0,3)\) respectively,

\(x+y=3\)

\(\begin{array}{|l|l|l|} \hline \mathrm{x} & 3 & 0 \\ \hline \mathrm{y} & 0 & 3 \\ \hline \end{array}\)

\(\mathrm{A}(3,0) ; \mathrm{B}(0,3)\)

Join points \(A\) and \(B\) to obtain the line.

Clearly, \((0,0)\) satisfies the in equation \(x+y \leq 3\).

So, the region containing the origin represent the solution set of the in equation.

Reglon represented by \(y \leq 6\) :

The line \(y=6\) is parallel to \(x\)-axis and its every point will satisfy the in equation in first, quadrant, region containing the origin represents the solution set of this in equation.

Reglon represented by \(x \geq 0\) and \(y \geq 0\) :

Since every point in first quadrant satisfy the in equations, so the first quadrant is the solution set of these in equations.

The shaded region is the common region of in equations. This is feasible region of solution which is bounded and is in first quadrant.